Finding local min, max, and saddle points in multivariable calculus

Find the local maximum and minimum values and saddle point(s) of the function.

f(x,y) = 1 + 2xy - x^2 - y^2

2. Relevant equations

The Second Derivative Test: let D = D(a,b) = fxx(a,b)*fyy(a,b) - [fxy(a,b)]^2
if D > 0 and fxx(a,b) > 0, then f(a,b) is a local minimum
if D > 0 and fxx(a,b) < 0, then f(a,b) is a local maximum
if D < 0 then f(a,b) is a saddle point
if D = 0 then the test is inconclusive

3. The attempt at a solution

I tried to use the Second Derivative Test to find the local mins, maxes, and saddle points but it's inconclusive, and I don't know how else to find them. My textbook says the answer is "f has a local maximum value of 1 at all points of the form (x, x)"

This is my work for the Second Derivative Test:

fx = 2y - 2x = 0 --> 2y = 2x --> y = x

fy = 2x - 2y = 0 --> 2x - 2(x) = 0 --> 0 = 0

so i guess there are critical points at every value where y = x... which matches the textbook's answer.
and then:

Rewriting the function as f(x, y) = (x - y)^2 + 1,
we see that the minimum value must be 1 (since 0 is the smallest value of a square),
and this is attained whenever y = x (i.e., points of the form (x, x)).

Staff: Mentor

Find the local maximum and minimum values and saddle point(s) of the function.

f(x,y) = 1 + 2xy - x^2 - y^2

2. Relevant equations

The Second Derivative Test: let D = D(a,b) = fxx(a,b)*fyy(a,b) - [fxy(a,b)]^2
if D > 0 and fxx(a,b) > 0, then f(a,b) is a local minimum
if D > 0 and fxx(a,b) < 0, then f(a,b) is a local maximum
if D < 0 then f(a,b) is a saddle point
if D = 0 then the test is inconclusive

3. The attempt at a solution

I tried to use the Second Derivative Test to find the local mins, maxes, and saddle points but it's inconclusive, and I don't know how else to find them. My textbook says the answer is "f has a local maximum value of 1 at all points of the form (x, x)"